\log_{10}{(x-3)} = \log{(x^2+15x-36)}+1 \\
\log{(x-3)} = \log{(x^2+15x-36)}+\log{10} \\
\log{(x-3)} = \log{(10\cdot(x^2+15x-36))} \\
x-3 = 10x^2+150x-360 \\
....... \\
x_1 = \frac{21}{10} = 2.1 \\
x_2 = -17